Question

A current of $0.5$ ampere when passed through $Ag{NO}_3$ solution for $193$ seconds deposited $0.108g$ of $Ag$. The equivalent weight of $Ag$ is:

• A. $10.8$
• B. $108$
• C. $54$
• D. $1$

Answer 1

The correct option is B $108$

Given: Current ,I = 0.5 A

Time,t = 193s

Deposited Ag (W) =0.108g

Solution:

Q=It , where Q= quantity of electricity in Coloumb, I is current in amperes and time is in seconds.

$\Rightarrow$Q=0.5 A× 193 s

$\Rightarrow$Q=96.5C

$AgN{O}_3\to A{g}^{+1}+N{O}_3^{-1}$

$A{g}^{+1}+e^{-}\to Ag$

Therefore, Number of electrons , n=1

96.5 C of electricity deposits Ag=0.108g

96500 C of electricity deposits Ag =$\frac{0.108}{96.5}\times 96500$

= 108g

Therefore, equivalent weight is E=$\frac{108}{n}$

$\Rightarrow$E= 108 g

Hence, option B is correct.